Determinants Matrices Question 96
Question: Matrix A such that $ A^{2}=2A-I $ , where I is the identity matrix, then for $ n\ge 2,{A^{n}} $ is equal to
Options:
A) $ {2^{n-1}}A-(n-1)I $
B) $ {2^{n-1}}A-I $
C) $ nA-(n-1)I $
D) $ nA-I $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Given $ A^{2}=2A-I $ Now $ A^{3}=A(A^{2})=A(2A-I) $
$ =2A^{2}-A=2(2A-I)-A=3A-2I $
$ A^{4}=A(A^{3})=A(3A-2I) $
$ =3A^{2}-2A=3(2A-I)-2A=4A-3I $ Following this, we can say $ A^{n}=nA-(n-I)I $
BETA
